# Thread: trigonometric

1. ## trigonometric

I have 2 questions:

1. find for z ; where z^3=sqrt[imaginary]
2. Simplify: sin x/(1-cos x)
can somebody help me with these 2 questions showing working out?
Please & Thank You

2. Hello, n7167113!

Find $\displaystyle z\!:\;\;z^3\:=\:\sqrt{i}$
We have: .$\displaystyle z \:=\:i^{\frac{1}{6}}$

We know: .$\displaystyle i \:=\:\cos\tfrac{\pi}{2} + i\sin\tfrac{\pi}{2}$,

Use DeMoivre's Formula for the six roots of: .$\displaystyle \left(\cos\tfrac{\pi}{2} + i\sin\tfrac{\pi}{2}\right)^{\frac{1}{6}}$

Simplify: .$\displaystyle \frac{\sin x}{1-\cos x}$
This doesn't simplify very much, but here goes . . .

Multiply by $\displaystyle \frac{1 +\cos x}{1+\cos x}\!:\;\;\frac{\sin x}{1 - \cos x}\cdot\frac{1+\cos x}{1+\cos x} \;=\;\frac{\sin x(1 + \cos x)}{1-\cos^2\!x} \;=\;\frac{\sin x(1 + \cos x)}{\sin^2\!x}$

. . $\displaystyle = \;\frac{1+\cos x}{\sin x} \;=\;\frac{1}{\sin x} + \frac{\cos x}{\sin x} \;=\;\csc x + \cot x$

3. For the simplification one, try multiplying both the numerator and denominator by (1+cos(x)), and from there you should be able to get a nice answer

edit> I was beat to the punch